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Therefore,
4z þ
6
:
4
H(z)
¼
z
2
þ
0
:
5z þ
0
:
6
4.8 SOLUTION OF LTI DISCRETE-TIME STATE EQUATIONS
Consider a discrete LTI system described by the state-space equation
x
(
k
þ
1
) ¼
Ax
(
k
) þ
Bu
(
k
)
(
4
:
100
)
and the output equation
y
(
k
) ¼
Cx
(
k
) þ
Du
(
k
)
(
:
)
4
101
To obtain the output of this system for a given input and an initial state, we
first solve
the state equation (Equation 4.100) and then the solution is substituted into the
algebraic equation (Equation 4.101) in order to
. The solution to
state equations has two parts, a homogeneous solution and a particular solution. We
nd the output y
(
k
)
first consider the homogeneous solution.
4.8.1 S
OLUTION OF
H
OMOGENEOUS
S
TATE
E
QUATION
The homogeneous solution is the solution of state equations to an arbitrary initial
condition with zero input. The homogeneous state equation is given by
x
(
k
þ
1
) ¼
Ax
(
k
)
(
4
:
102
)
Assuming that the initial state is x
(
0
)
, we have
x
(
1
) ¼
Ax
(
0
)
) ¼
A
2
x
(
x
(
2
) ¼
Ax
(
1
) ¼
A
Ax
(
0
0
)
) ¼
A
A
2
x
(
) ¼
A
3
x
(
x
(
3
) ¼
Ax
(
2
0
0
)
(
4
:
103
)
.
x
(
k
) ¼
Ax
(
k
) ¼
A
A
k
1
x
(
) ¼
A
k
x
(
1
0
0
)
Therefore, in matrix form, the solution to the homogeneous equation is
x
(
k
) ¼
A
k
x
(
0
)
(
4
:
104
)
Therefore, the homogeneous part of the solution is x
(
k
) ¼
A
k
x
(
0
)
. The matrix expo-
nential A
k
is called state-transition matrix and is denoted by
w(
k
)
. The state-transition
matrix
w(
k
)
is an N
N matrix and is the solution to the homogeneous equation
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